[树 🌲] #2

Linjiayu6 · 2020-06-07T01:44:12Z

No description provided.

Linjiayu6 · 2020-06-07T02:32:41Z

0 - 知识点

DFS: 深度优先搜索

前序遍历 root => left => right
中序遍历 left => root => right
后序遍历 left => right => root

BFS: 广度优先搜索

方法: 递归迭代

Linjiayu6 · 2020-06-07T03:56:53Z

1 - 先序遍历

144. 二叉树的前序遍历

589. N叉树的前序遍历

114. 二叉树展开为链表给定一个二叉树，原地将它展开为一个单链表

方法1 - 迭代入栈

root 先入栈执行, 虽然是left 优先级大于right, 但因为是入栈 pop出去的是栈顶内容, 所以先right 入栈, 后left 入栈, 再left出栈, right出栈。

# 迭代
class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        stack, result = [root], []
        
        while stack: # 栈一直有值
            root = stack.pop()
            # root优先级最高
            if root is not None: 
                result.append(root.val)
                # 第二优先级应该是left, 但因为是数组, 最先pop的是优先级高的, 所以是right先入栈
                if root.right is not None: stack.append(root.right)
                if root.left is not None: stack.append(root.left)
                
        return result

var preorderTraversal = function(root) {
    if (!root) {
        return [];
    }

    var result = [];
    var stack = [root];

    while (stack.length > 0) {
        var node = stack.pop();
        result.push(node.val);
        if (node) {
            node.right && stack.push(node.right);
            node.left && stack.push(node.left);
        }
    }

    return result;
};

方法2 - 递归分裂

class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        def traverse(root):
            if root is None: return [] # 必须给这个结束标识
            return [root.val] + traverse(root.left) + traverse(root.right)
        return traverse(root)

// JS  递归
var preorderTraversal = function(root) {
    var result = [];

    function traverse (node, result) {
        if (node) {
            result.push(node.val);
            traverse(node.left, result);
            traverse(node.right, result);
        }
    }

    traverse(root, result);
    return result;
};

# 114. 二叉树展开为链表
class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        # 深度优先搜索 前序遍历 中左右
        if root is None: return None

        stack = [(root, 'light')]
        result = TreeNode(None)
        tmp = result
        while stack:
            (node, flag) = stack.pop()
            if node:
                if flag == 'light':
                    if node.right: stack.append((node.right, 'light'))
                    if node.left: stack.append((node.left, 'light'))
                    node.left = None # 把自己变成叶子节点
                    node.right = None
                    stack.append((node, 'grey'))
                else:
                    tmp.right = node
                    tmp = tmp.right
        return result.right

Linjiayu6 · 2020-06-07T08:03:24Z

2 - 中序遍历

94. 二叉树的中序遍历

173. 二叉搜索树迭代器

迭代 1

思路:
root还是会做为初始化节点输入进去
1. root pop出去，检查不是叶子🍃节点 / 是否有左右子树，就将 right => root => left 一次入栈。此时设置root是置灰状态。说明下次遇到它的时候，不会去检查是否有left和right, 而是直接返回值。
2. 如果不是置灰颜色的，就去检查是否有左子树右子树。
......

# 迭代
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        if root is None: return []

        stack, result = [(root, 'active')], []
        while stack:
            (node, status) = stack.pop()

            if status == 'active':
                node.right and stack.append((node.right, 'active')) # right后执行, 先入栈
                stack.append((node, 'grey'))
                node.left and stack.append((node.left, 'active')) # left最先执行, 后入栈
            else:
                result.append(node.val) # 置灰状态直接返回值, 说明已经遍历完左子树了。
        return result

// 这个写法并不是非常的好
var inorderTraversal = function(root) {
    var result = [];
    var stack = [{ node: root, isTraversed: false }];

    while (stack.length > 0) {
        var { node, isTraversed } = stack.pop();

        if (node) {
            if (isTraversed === true) {
                result.push(node.val);
            } else {
                node.right && stack.push({ node: node.right, isTraversed: false });
                stack.push({ node, isTraversed: true });
                node.left && stack.push({ node: node.left, isTraversed: false });
            }
        }
    }

    return result;
};

迭代 2

核心是 [左，中，右]，将左侧作为中心位置，依次 node.left 放入栈中，直到叶子节点，后pop，再将右侧放入栈中

var inorderTraversal = function(root) {
    if (!root) return [];

    var result = [];
    var stack = [];

    while (root || stack.length > 0) {
        // 左侧子树, 全部放进栈中
        while (root) {
            stack.push(root);
            root = root.left;
        }

        var node = stack.pop();
        result.push(node.val);
       // 遇到右侧子树，作为新节点，再次左侧子树处理
        root = node.right;
    }
    return result;
};

递归

# 递归
class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        # left => root => right
        def traverse(node):
            if node is None: return []
            return traverse(node.left) + [node.val] + traverse(node.right)
        
        return traverse(root)

var inorderTraversal = function(root) {
    var result = [];

    function traverse (node, result) {
        if (node) {
            traverse(node.left, result);
            result.push(node.val);
            traverse(node.right, result);
        }
    }

    traverse(root, result);

    return result;
};

Linjiayu6 · 2020-06-07T08:28:54Z

3 - 后序遍历

145. 二叉树的后序遍历

# 迭代 同上面一样, 取色 取状态判断
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        # 迭代
        if root is None: return []
        stack, result = [(root, 'active')], []
        while stack:
            (node, status) = stack.pop()
            if status == 'active':
                stack.append((node, 'grey'))
                node.right and stack.append((node.right, 'active'))
                node.left and stack.append((node.left, 'active'))
            else:
                result.append(node.val)
        return result

var postorderTraversal = function(root) {
    if (!root) return [];
    var stack = [{ node: root, flag: false }];
    var result = [];

    while (stack.length > 0) {
        var { node, flag } = stack.pop();
        if (flag === true) {
            result.push(node.val);
        } else {
            // [中, 右, 左]
            node && stack.push({ node, flag: true });
            node.right && stack.push({ node: node.right, flag: false });
            node.left && stack.push({ node: node.left, flag: false });
        }
    }

    return result;
};

# 递归
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        # 递归
        if root is None: return []
        return self.postorderTraversal(root.left) + self.postorderTraversal(root.right) + [root.val]

var postorderTraversal = function(root) {
    var result = [];
    
    function traverse (node, result) {
        if (node) {
            traverse(node.left, result);
            traverse(node.right, result);
            result.push(node.val);
        }
    }
    traverse(root, result);
    return result;
};

剑指 Offer 54. 二叉搜索树的第k大节点

通过遍历树来解决, 优先级为右中左

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def kthLargest(self, root: TreeNode, k: int) -> int:
        # right root left遍历
        if root is None: return root
        stack = [(root, 'light')]
        x = 1
        while stack:
            (node, flag) = stack.pop() # pop最后一个
            if node:
                if flag == 'light':
                    # 右中左, 后执行后入栈
                    if node.left: stack.append((node.left, 'light'))
                    stack.append((node, 'grey'))
                    if node.right: stack.append((node.right, 'light'))
                else:
                    if x != k:  x += 1
                    else: return node.val

Linjiayu6 · 2020-06-07T10:16:53Z

4 - 广度优先搜索

102. 二叉树的层序遍历

116. 填充每个节点的下一个右侧节点指针

117. 填充每个节点的下一个右侧节点指针 II

# 迭代
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        """
        current = [当前执行节点]
        child = [当前执行节点.left, 当前执行节点.right]
        遍历current, 当current遍历结束, 则current = child, child = []
        直到current为空, 结束
        """
        if root is None: return []
        current = [root]
        result = []
        while current:
            tmp, child = [], []
            for node in current:
                tmp.append(node.val)
                if node.left: child.append(node.left)
                if node.right: child.append(node.right)
            result.append(tmp)

            current = child
        return result

// 层次遍历 递归
var levelOrder = function(root) {
    if (!root) return []
    var result = []
    
    var traverse = (node, i) => {
        if (!node) return
        if (!result[i] || result[i] && result[i].length === 0) {
            result[i] = [node.val]
        } else {
            result[i].push(node.val)
        }

        return traverse(node.left, i + 1) || traverse(node.right, i + 1)
    }

    traverse(root, 0)
    return result
}

// 递归
var levelOrder = function(root) {
    if (!root) return [];

    var result = [];
    function traverse (node, layer = 0) {
        if (node) {
            if (result[layer]) {
                result[layer].push(node.val);
            } else {
                result.push([node.val]);
            }
        }

        node.left && traverse(node.left, layer + 1);
        node.right && traverse(node.right, layer + 1);
    }


    traverse(root, 0);
    return result;
};

// 迭代
var levelOrder = function(root) {
    if (!root) return [];

    var result = [];
    var cur_layer = [root];
    var child_layer = [];

    while (cur_layer.length > 0) {
        var tmp_result = [];
        for (i = 0; i < cur_layer.length; i++) {
            var node = cur_layer[i];
            if (node) {
                tmp_result.push(node.val);
                node.left && child_layer.push(node.left);
                node.right && child_layer.push(node.right);
            }
        }

        cur_layer = child_layer;
        child_layer = [];
        result.push(tmp_result);
    }
    
    return result;
};

Linjiayu6 · 2020-06-09T02:42:23Z

5 - 最大深度

104. 二叉树的最大深度

559. N叉树的最大深度

广度优先搜索: 最后返回的是数组长度, 同 102. 二叉树的层序遍历
深度优先搜索: 返回的是max(left, right)
递归说明

# 递归
class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if root is None: return 0
        return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1

# 迭代
class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        # 先序遍历, 每层深度保存, 并和暂存区最大对比
        if root is None: return 0
        
        stack = [(root, 1)] # 此时root是一个节点, 深度为1
        _max = 1

        while stack:
            (node, depth) = stack.pop()
            _max = max(_max, depth) # 当前深度和最大比较
            if node.right: stack.append((node.right, depth + 1)) # 先将右子树入栈, 并深度 +1
            if node.left: stack.append((node.left, depth + 1)) # 先将左子树入栈, 并深度 +1
        return _max

Linjiayu6 · 2020-06-09T02:42:41Z

6 - * 对称二叉树 * 未独立思考出来 TODO 迭代

101. 对称二叉树

镜子原理, 整出来两棵树, p1 和 p2 进行对比对比逻辑是
- p1.left == p2.right
- p1.right== p2.left

# 递归
class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        # 镜子
        def mirror(p1, p2):
            if p1 is None and p2 is None: # 都是空, 则返回True
                return True
            if p1 and p2 and p1.val == p2.val: # 当前结点 值 相同, 则去比较子树内容。
                return mirror(p1.left, p2.right) and mirror(p1.right, p2.left)
            return False

        return mirror(root, root)

# 递归 在一棵树上的递归遍历
class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        if root is None: return True

        def compare(p1, p2):
            if p1 is None and p2 is None: return True # 叶子节点
            if p1 and p2 and p1.val == p2.val: # 那就继续走下去
                return compare(p1.left, p2.right) and compare(p1.right, p2.left)
            return False
        return compare(root.left, root.right)

Linjiayu6 · 2020-06-09T02:42:56Z

7 - 路径总和

112. 路径总和

广度优先遍历, 值 & 为叶子节点即返回true

# 递归
class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if root is None:
            return False

        if root.left is None and root.right is None:
            return root.val == sum  # 是个叶子节点, 且值匹配

        sum -= root.val
        return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)

# 迭代
class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        # 深度优先搜索, 中序遍历
        # 广度优先搜索
        if root is None: return False

        current = [root]
        while current:
            child = []
            for i in current:
               # 如果到头了 是个叶子节点, 就比较值 有的话就直接返回
                if i.val == sum and i.left is None and i.right is None: 
                    return True
                if i.left:
                    i.left.val += i.val
                    child.append(i.left)

                if i.right:
                    i.right.val += i.val
                    child.append(i.right)
            
            current = child
        return False

剑指 Offer 34. 二叉树中和为某一值的路径进阶🔥

同 113. 路径总和 II

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        if root is None: return []
        if root and root.left is None and root.right is None:
            return [[root.val]] if root.val == sum else []
       
        result = []
        def traverse (node, _sum, tmpList):
            if node:
                val = node.val
                # 叶子节点 且 值为0
                if _sum - val == 0 and node.left is None and node.right is None:
                    result.append(tmpList + [val])
                
                traverse(node.left, _sum - val, tmpList + [val])
                traverse(node.right, _sum - val, tmpList + [val])

        traverse(root.left, sum - root.val, [root.val])
        traverse(root.right, sum - root.val, [root.val])
        return result

Linjiayu6 · 2020-06-09T11:58:10Z

8 - 二叉树镜像

面试题27. 二叉树的镜像

同 226. 翻转二叉树

# 递归
class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        if root is None: return None
        def swap(node):
            if node:
                node.left, node.right = node.right, node.left # 交换
                swap(node.left) # 继续处理左子树
                swap(node.right) # 继续处理右子树
        swap(root)
        return root

# 迭代
class Solution(object):
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        # 迭代 广度优先遍历, 将节点遍历交换
        if root is None: return None
        stack = [root]
        while stack:
            child = [] # 将节点放到child里面
            for node in stack:
                if node:
                    node.left, node.right = node.right, node.left # 将节点left, right 交换
                    child.append(node.left)
                    child.append(node.right)
            stack = child
        return root

Linjiayu6 · 2020-06-09T12:15:16Z

9 - 树的子结构

面试题26. 树的子结构

# 广度优先迭代  +  isSubTree 子节点判断递归

class Solution:
    def isSubTree(self, A, B):
        if A is None and B: return False # A都遍历完了, B还有
        if B is None: return True # B反正遍历完了, 返回True

        # 是否是子树结构
        if A.val == B.val:
            return self.isSubTree(A.left, B.left) and self.isSubTree(A.right, B.right)

    def isSubStructure(self, A: TreeNode, B: TreeNode) -> bool:
        # 广度优先遍历
        if A is None or B is None: return False
        stack = [A]
        while stack:
            child = []
            for node in stack:
                if node:
                    if node.val == B.val:  # 如果节点相同，就去找
                        if self.isSubTree(node, B) == True: return True # 判断当前结点是否是子节点
                    child.append(node.left) # 继续找
                    child.append(node.right) 

            stack = child
        return False

Linjiayu6 · 2020-06-19T11:41:29Z

10 - 重建二叉树

105. 从前序与中序遍历序列构造二叉树

同剑指 Offer 07. 重建二叉树

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
        if len(preorder) == 0: return None
        if len(preorder) == 1: return TreeNode(preorder[0])
        
        root_val = preorder.pop(0) # 先序遍历第一个肯定是 root
        root = TreeNode(root_val) # 创建root

        # i = 0
        # while i < len(inorder):
        #     if inorder[i] == root_val: 
        #         break
        #     i+= 1
        i = inorder.index(root_val) # 找位置

        root.left = self.buildTree(preorder[0: i], inorder[0: i])
        root.right = self.buildTree(preorder[i: ], inorder[i+1: ])
        return root

106. 从中序与后序遍历序列构造二叉树

# 跟上面一样的类型，思路要整理清晰
class Solution(object):
    def buildTree(self, inorder, postorder):
        if len(inorder) == 0: return None
        if len(inorder) == 1: return TreeNode(inorder[0])
        # postorder[-1] 最后一个肯定是 root

        root_val = postorder[-1]
        root = TreeNode(root_val)
        index = inorder.index(root_val) # 找值, 返回 index
        
        root.left = self.buildTree(inorder[0: index], postorder[0: index])
        root.right = self.buildTree(inorder[index + 1: ], postorder[index: len(postorder) - 1])
        return root

Linjiayu6 · 2020-06-20T02:52:38Z

11 - 二叉树🌲 公共祖先 / 二叉搜索树🌲 公共祖先

236. 二叉树的最近公共祖先

同剑指 Offer 68 - II. 二叉树的最近公共祖先

二叉树🌲 公共祖先

class Solution:
    def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
        if root is None: return None
        
        # 有一个是root点, 就返回root
        if root == p or root == q: return root
        # 左右子树在同一层上, 就返回root
        if root.left == p and root.right == q or root.right == p and root.left == q: return root
        
        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)
        if left is None or right is None: return right or left
        else: return root

二叉搜索树🌲 公共祖先

剑指 Offer 68 - I. 二叉搜索树的最近公共祖先

同 235. 二叉搜索树的最近公共祖先

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if root is None: return None
        if root == p or root == q: return root
        if (root.left == p and root.right == q) or (root.left == q and root.right == p): return root

        # 都在左侧
        if p.val < root.val and q.val < root.val: return self.lowestCommonAncestor(root.left, p, q)
        # 都在右侧
        if p.val > root.val and q.val > root.val: return self.lowestCommonAncestor(root.right, p, q)
        # 在左右侧都有, 则返回root
        return root

Linjiayu6 mentioned this issue Jul 26, 2020

12 - 算法 II Linjiayu6/FE-Notes#12

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[树 🌲] #2

[树 🌲] #2

Linjiayu6 commented Jun 7, 2020

Linjiayu6 commented Jun 7, 2020 •

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Linjiayu6 commented Jun 7, 2020 •

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Linjiayu6 commented Jun 7, 2020 •

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Linjiayu6 commented Jun 7, 2020 •

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Linjiayu6 commented Jun 7, 2020 •

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Linjiayu6 commented Jun 9, 2020 •

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Linjiayu6 commented Jun 9, 2020 •

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Linjiayu6 commented Jun 9, 2020 •

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Linjiayu6 commented Jun 9, 2020

Linjiayu6 commented Jun 9, 2020 •

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Linjiayu6 commented Jun 19, 2020 •

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Linjiayu6 commented Jun 20, 2020 •

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[树 🌲] #2

[树 🌲] #2

Comments

Linjiayu6 commented Jun 7, 2020

Linjiayu6 commented Jun 7, 2020 • edited Loading

0 - 知识点

DFS: 深度优先搜索

BFS: 广度优先搜索

方法: 递归 迭代

Linjiayu6 commented Jun 7, 2020 • edited Loading

1 - 先序遍历

114. 二叉树展开为链表 给定一个二叉树，原地将它展开为一个单链表

方法1 - 迭代 入栈

方法2 - 递归 分裂

Linjiayu6 commented Jun 7, 2020 • edited Loading

2 - 中序遍历

迭代 1

迭代 2

递归

Linjiayu6 commented Jun 7, 2020 • edited Loading

3 - 后序遍历

Linjiayu6 commented Jun 7, 2020 • edited Loading

4 - 广度优先搜索

Linjiayu6 commented Jun 9, 2020 • edited Loading

5 - 最大深度

Linjiayu6 commented Jun 9, 2020 • edited Loading

6 - * 对称二叉树 * 未独立思考出来 TODO 迭代

Linjiayu6 commented Jun 9, 2020 • edited Loading

7 - 路径总和

剑指 Offer 34. 二叉树中和为某一值的路径 进阶🔥

同 113. 路径总和 II

Linjiayu6 commented Jun 9, 2020

8 - 二叉树镜像

同 226. 翻转二叉树

Linjiayu6 commented Jun 9, 2020 • edited Loading

9 - 树的子结构

Linjiayu6 commented Jun 19, 2020 • edited Loading

10 - 重建二叉树

同 剑指 Offer 07. 重建二叉树

Linjiayu6 commented Jun 20, 2020 • edited Loading

11 - 二叉树🌲 公共祖先 / 二叉搜索树🌲 公共祖先

同 剑指 Offer 68 - II. 二叉树的最近公共祖先

二叉树🌲 公共祖先

二叉搜索树🌲 公共祖先

同 235. 二叉搜索树的最近公共祖先

Linjiayu6 commented Jun 7, 2020 •

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方法: 递归迭代

Linjiayu6 commented Jun 7, 2020 •

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114. 二叉树展开为链表给定一个二叉树，原地将它展开为一个单链表

方法1 - 迭代入栈

方法2 - 递归分裂

Linjiayu6 commented Jun 7, 2020 •

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Linjiayu6 commented Jun 7, 2020 •

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Linjiayu6 commented Jun 7, 2020 •

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Linjiayu6 commented Jun 9, 2020 •

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Linjiayu6 commented Jun 9, 2020 •

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Linjiayu6 commented Jun 9, 2020 •

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剑指 Offer 34. 二叉树中和为某一值的路径进阶🔥

Linjiayu6 commented Jun 9, 2020 •

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Linjiayu6 commented Jun 19, 2020 •

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同剑指 Offer 07. 重建二叉树

Linjiayu6 commented Jun 20, 2020 •

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同剑指 Offer 68 - II. 二叉树的最近公共祖先